Monty Hall Problem: Maximize Your Chances of Winning a Car

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In summary, the Monty Hall problem is a probability puzzle where a contestant is presented with three doors, one of which has a prize behind it. After choosing a door, the host reveals one of the other doors that does not have the prize. The optimal strategy for this problem is to always switch your choice to the remaining door, as this increases your chances of winning. This is due to the concept of conditional probability, where the likelihood of your initial choice being wrong is higher than it being correct. This problem also has real-world applications in decision-making and risk assessment. While there are variations of the Monty Hall problem, the optimal strategy remains the same.
  • #1
hatelove
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There are three doors and one of the doors with a goat are removed. Which of the two remaining doors should you choose to have a higher chance of getting the car?

There are ten doors and eight of the doors with goats are removed. Which of the two remaining doors should you choose to have a higher chance of getting the car?

There are two doors. Which of the two doors should you choose to have a higher chance of getting the car?

Are all of these the exact same version of the Monty Hall problem?
If so, I don't understand why it's not 50/50 in all cases.
If not, I don't understand how these are any different from the Monty Hall problem.
 
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  • #2
daigo said:
Are all of these the exact same version of the Monty Hall problem?
If so, I don't understand why it's not 50/50 in all cases.
If not, I don't understand how these are any different from the Monty Hall problem.

Please post the entire question

CB
 
  • #3
Which entire question? I am just wanting to compare the aforementioned 3 scenarios to the Monty Hall/3 Prisoners/Bertrand's Box problem
 
  • #4
If these quotes contain the whole problem, they are different from the Monty Hall problem because:

1. you do not begin by choosing a door (which cannot be reveailed in the next step)
2. the algorithm of the door-opening is not given (could it have relealed a car?) <- not relevant here due to 1, but important in Monty Hall.
 
  • #5
Why does it make a difference if you begin by choosing a door? In the end, don't you only have two choices either way in both the quoted problems and the Monty Hall? (Door #1 or Door #2)

Also I'm not sure what you mean by your second point.
 
  • #6
It makes a huge difference if you choose before and are asked to switch, which is the set up of the Monty Hall problem. The best way to understand the Monty Hall problem is there are two ways to switch:

1) Correct door to incorrect door
2) Incorrect door to correct door

You cannot go incorrect to incorrect or correct to correct. So the question is how often do you choose a correct door? Well that is just 1/3 of the time. How often do you choose an incorrect door? 2/3 of the time. So if you always switch, 2/3 of the time you will switch to the correct door.

If you are just given two doors and the prize is behind one of them, then your guess is 50/50. The Monty Hall situation is different than that.
 
  • #7
daigo said:
Why does it make a difference if you begin by choosing a door?
It does influence Mr. Monty Hall, who has to open an empty door, but may not use your door.
 
  • #8

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  • #9
I love this problem for yet another reason now. :)

Earlier this year in one of my classes on probability, this problem was introduced and it was fun as always to watch people get stuck and even disagree with the solution very stubbornly. So I suppose this thread is a good place to write down some more comments as it pertains to the OP still.

If I were teaching this concept, I would do two things to help convince the students:

1) Run a test program with a large number of iterations to show that switching converges to a 66.7% success rate.

2) Reword the problem such that instead of going from 3 doors to 2 doors, go from 1,000,000 doors to 2 doors. Now it should become very obvious to most everyone that the door they original chose is almost never the correct door.
 
  • #10
Jameson said:
...
2) Reword the problem such that instead of going from 3 doors to 2 doors, go from 1,000,000 doors to 2 doors. Now it should become very obvious to most everyone that the door they original chose is almost never the correct door.

I have used this tactic with success in converting "nonbelievers." :D
 
  • #11
It's a very nice modification to the problem to account for our very poor statistical intuition. Even so, I think the answer can be so puzzling and counter-intuitive to some that this example still won't convince them. That's why I suggested the computer simulation, which confirms that the theory at work isn't some hypothetical mind trick.
 

FAQ: Monty Hall Problem: Maximize Your Chances of Winning a Car

What is the Monty Hall problem?

The Monty Hall problem is a probability puzzle named after the host of the popular game show "Let's Make a Deal", Monty Hall. In the problem, you are presented with three doors, one of which has a car behind it. After you choose a door, Monty reveals one of the other doors that does not have a car behind it. He then gives you the option to switch your choice to the remaining door. The question is, should you switch or stick with your original choice?

What is the optimal strategy for the Monty Hall problem?

The optimal strategy for the Monty Hall problem is to always switch your choice to the remaining door after Monty reveals one of the other doors. This strategy gives you a higher chance of winning the car, as it takes into account the probability of Monty's actions and the fact that your initial choice had a lower probability of being correct.

Why is the optimal strategy to switch in the Monty Hall problem?

The optimal strategy to switch in the Monty Hall problem is based on the concept of conditional probability. When you initially choose a door, there is a 1/3 chance that you have chosen the correct door. However, after Monty reveals one of the other doors, there is a 2/3 chance that the car is behind the remaining door. By switching, you are essentially betting on the fact that your initial choice was more likely to be wrong.

Does the Monty Hall problem have any real-world applications?

Although the Monty Hall problem is a famous puzzle, it does have real-world applications in areas such as decision-making and risk assessment. The problem helps to illustrate the importance of taking into account all available information and considering the probability of different outcomes when making decisions.

Are there any variations of the Monty Hall problem?

Yes, there are several variations of the Monty Hall problem that involve different numbers of doors or different rules for Monty's actions. However, the underlying concept of conditional probability remains the same. The optimal strategy for these variations may differ slightly, but it is always best to switch your choice if given the option.

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